Consider $f: \mathbf{R}_{+} \rightarrow[4, \infty)$ given by $f(x)=x^{2}+4$. Show that $f$ is invertible with the inverse $f^{-1}$ of given $f$ by $f^{-1}(y)=\sqrt{y-4}$ where $\mathbf{R}_{+}$is the set of all non-negative real numbers.

Solution:

$ f: R_+ → [4, ∞) is \ given\ as f(x) = x^2 + 4. \\ $

One-one:

Let, $$ f(x)=f(y) \\ $$

$$ \begin{aligned} &\Rightarrow x^{2}+4=y^{2}+4 \\\\ &\Rightarrow x^{2}=y^{2} \\\\ &\Rightarrow x=y \quad\left[\text { as } x=y \in \mathbf{R}_{+}\right] \\ \end{aligned} $$

$\therefore f$ is a one-one function.

Onto:

For , $ y \in[4, \infty) \\ $

let , $$ y=x^{2}+4 \\ $$

$$ \begin{aligned} &\Rightarrow x^{2}=y-4 \geq 0 \quad[\text { as } y \geq 4] \\\\ &\Rightarrow x=\sqrt{y-4} \geq 0 \\ \end{aligned} $$

Therefore, for any $y \in \mathbf{R}$, there exists $x=\sqrt{y-4} \in \mathbf{R}_{\text {such that }}$

$$ f(x)=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=y-4+4=y . \\ $$

$\therefore f$, is onto.

Thus, f is one-one and onto and therefore, $f^{-1}$ exists.

Let us define,

$$ g:[4, \infty) \rightarrow \mathbf{R}_{+} \\ $$

$$ g(y)=\sqrt{y-4} \\ $$

Now,

$$ g \circ f(x)=g(f(x))=g\left(x^{2}+4\right)=\sqrt{\left(x^{2}+4\right)-4}=\sqrt{x^{2}}=x \\ $$

And,

$$ f \circ g(y)=f(g(y))=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=(y-4)+4=y \\ $$

$$ \therefore g \circ f=f \circ g=\mathrm{I}_{\mathrm{R}+} \\ $$

Hence, f is invertible and the inverse of $f$ is given by, $$ f^{-1}(y)=g(y)=\sqrt{y-4} \text {. } \ $$